# Download A second course in general topology by Heikki Junnila PDF By Heikki Junnila

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By Lemma 4, there exists a continuous mapping fn : X → Y such that we have d(fn (x), θ(x)) < 2−n−1 for every x ∈ X. The definition of θ shows that fn satisfies both required conditions. This completes the induction. By completeness of Y , there exists a function f : X → Y such that fn (x) → f (x) for every x ∈ X. The convergence is uniform, and hence the mapping f is continuous. For every x ∈ X, we have that d(f (x), ϕ(x)) = 0, and it follows, since ϕ(x) ⊂c Y , that f (x) ∈ ϕ(x). As a consequence, f is a continuous selection of ϕ.

We can define a mapping ϕ : X → Sℓ2A by the formula ϕ(x)α = fα (x). We show that ϕ is an embedding. Since X is T1 and F induces the topology of X, we see that ϕ is one-to-one. To show that ϕ is continuous, let xn → x in X. We show that ϕ(xn ) → ϕ(x), in other words, that √ ||ϕ(xn ) − ϕ(x)|| → 0. For all a, b ≥ 0 we have that ab ≤ 12 (a + b), and it follows that we have, for every n ∈ N that α∈A fα (xn )fα (x) ≤ 1 2 fα (xn ) + α∈A and hence that 49 α∈A fα (x) = 1 < ∞ ||ϕ(xn ) − ϕ(x)||2 = fα (xn ) + α∈A α∈A fα (x) − α∈A fα (xn ) − α∈A fα (x) 2 fα (xn )fα (x) = 2 − 2 2 = fα (xn )fα (x) .

Proof. Let g : X By Proposition 11, f is continuous, and it follows by Lemma 17, that f is Z-uniformly ˇ continuous. By Lemma 16, there exists an Z-uniformly continuous function fˇ : UZ → R such that we have fˇ(Zx ) = f (x) for every x ∈ X. The function fˇ is an extension of g and Lemma 15 shows that fˇ is continuous UZ → I. ˜ and we have another representation of the Cech-Stone ˇ We can thus write UZ = β X, compactification of X. , with discrete spaces. 19 Example Let D be a discrete space. Then every subset of D is a zero-set, that is, ZD = P(D).